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PilotLPTM [1.2K]
3 years ago
7

The rate constant of a reaction is 5.8 × 10−3 s−1 at 25°c, and the activation energy is 33.6 kj/mol. what is k at 75°c? enter yo

ur answer in scientific notation.
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

k₂ = 4.06 x 10⁻² s⁻¹.

Explanation:

  • From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

  • At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₁ = 5.8 × 10⁻³ s⁻¹, k₂ = ??? , Ea = 33600 J/mol, R = 8.314 J/mol.K, T₁ = 298.0 K, T₂ = 348.0 K.

  • ln(k₂/5.8 × 10⁻³ s⁻¹) = (33600 J/mol / 8.314 J/mol.K) [(348.0 K - 298.0 K) / (298.0 K x 348.0 K)] = (4041.37) (4.82 x 10⁻⁴) = 1.9479.
  • Taking exponential of both sides:

(k₂/5.8 × 10⁻³ s⁻¹) = 7.014.

∴ k₂ = 4.06 x 10⁻² s⁻¹.

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A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t
Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 2 atm
  • T1= 50 C= 323 K (being 0 C= 273 K)
  • P2= 3.2 atm
  • T2= ?

Replacing:

\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}

Solving:

T2*\frac{2 atm}{323 K} =3.2 atm

T2=3.2 atm*\frac{323 K}{2 atm}

T2= 516.8 K= 243.8 C

<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

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3 years ago
Bromine is is reddish-brown liquid that boils at 59 degrees. bromine is highly reactive with many metals, for example, it reacts
katrin2010 [14]
A physical property of an element is a property of an element that can observed or measured without changing the chemical nature of the element.

A chemical property of an element is a property of an element that can only be observed or measure when the chemical property of the element is altered or changed.

Based on this;
The boiling point of bromine is a physical property of bromine.
The high reactivity of bromine with many elements is a chemical property of bromine.
4 0
3 years ago
How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?
ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

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5 0
3 years ago
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Calculate the pH of a solution prepared by dissolving 0.150 mol of benzoic acid and 0.300 mol of sodium benzoate in water suffic
jok3333 [9.3K]

Answer : The pH of a solution is, 4.50

Explanation : Given,

K_a=6.30\times 10^{-5}

Concentration of benzoic acid (Acid) = 0.150 M

Concentration of sodium benzoate (salt) = 0.300 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.30\times 10^{-5})

pK_a=5-\log (6.30)

pK_a=4.20

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=4.20+\log (\frac{0.300}{0.150})

pH=4.50

Thus, the pH of a solution is, 4.50

8 0
3 years ago
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